Lemma 10.12.10 (00DF)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.12: Tensor products
Lemma 10.12.10 (cite)

numberstags

$\begin{align*} M_1\xrightarrow {f} M_2\xrightarrow {g} M_3 \to 0 \end{align*}$

be an exact sequence of $R$ -modules and homomorphisms, and let $N$ be any $R$ -module. Then the sequence

10.12.10.1

$\begin{equation} \label{algebra-equation-2ndex} M_1\otimes N\xrightarrow {f \otimes 1} M_2\otimes N \xrightarrow {g \otimes 1} M_3\otimes N \to 0 \end{equation}$

is exact. In other words, the functor $- \otimes _ R N$ is right exact, in the sense that tensoring each term in the original right exact sequence preserves the exactness.

Proof. We apply the functor $\mathop{\mathrm{Hom}}\nolimits (-, \mathop{\mathrm{Hom}}\nolimits (N, P))$ to the first exact sequence. We obtain

$0 \to \mathop{\mathrm{Hom}}\nolimits (M_3, \mathop{\mathrm{Hom}}\nolimits (N, P)) \to \mathop{\mathrm{Hom}}\nolimits (M_2, \mathop{\mathrm{Hom}}\nolimits (N, P)) \to \mathop{\mathrm{Hom}}\nolimits (M_1, \mathop{\mathrm{Hom}}\nolimits (N, P))$

By Lemma 10.12.8, we have

$0 \to \mathop{\mathrm{Hom}}\nolimits (M_3 \otimes N, P) \to \mathop{\mathrm{Hom}}\nolimits (M_2 \otimes N, P) \to \mathop{\mathrm{Hom}}\nolimits (M_1 \otimes N, P)$

Using the pullback property again, we arrive at the desired exact sequence. $\square$

Comments (0)

There are also:

8 comment(s) on Section 10.12: Tensor products

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$ ). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Comments (0)

Post a comment